Let $u:\mathbb R \to \mathbb R$ and $C \in (0, \infty)$ such that $$ |u(x) - u(y)| \le C|x-y| \text{ for a.e. } x,y \in \mathbb R, $$ or more precisely there is a Lebesgue null subset $N$ of $\mathbb R$ such that $$ |u(x) - u(y)| \le C|x-y|, \quad \forall x,y \in N^c := \mathbb R \setminus N. $$
To make clear the proof of Brezis' Proposition 8.4, I would like to verify that
Lemma For all $h>0$, $$ |u(x+h) - u(x)| \le Ch \text{ for a.e. } x \in \mathbb R. $$
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
Let $\lambda$ be the Lebesgue measure on $\mathbb R$. It suffices to prove that $x+h \in N^c$ for $\lambda$-a.e. $x \in \mathbb R$. Consider the integral $$ I := \int_{\mathbb R} 1_{N} (x+h) \, \mathrm d x = \int_{\mathbb R} 1_{N-h} (x) \, \mathrm d x = \lambda(N-h). $$
Because $N$ is a $\lambda$-null set, so is $N-h$. Then $I=0$ and thus $1_{N} (x+h)=0$ for $\lambda$-a.e. $x \in \mathbb R$. Then $x+h \in N^c$ for $\lambda$-a.e. $x \in \mathbb R$. This completes the proof.
Your proof is OK.
Maybe an easier proof is to show that $$A:=\{x\in\mathbb R: x\in N \text{ or } x+h\in N\}$$ has measure zero, hence for $\lambda\text{-a.e. }x\in\mathbb R$ we have $x\in N^c$ and $x+h\in N^c$, and thus $|u(x+h)-u(x)|\leq Ch$.
To show that $\lambda(A)=0$, just note that $A= N\cup (N-h)$, hence $$\lambda(A)\leq\lambda(N)+\lambda(N-h)=0+0=0.$$